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Questions tagged [differential-geometry]

Differential geometry is the application of differential calculus in the setting of smooth manifolds (curves, surfaces and higher dimensional examples). Modern differential geometry focuses on "geometric structures" on such manifolds, such as bundles and connections; for questions not concerning such structures, use (differential-topology) instead. Use (symplectic-geometry), (riemannian-geometry), (complex-geometry), or (lie-groups) when more appropriate.

Differential geometry is a branch of mathematics that uses techniques in calculus and linear algebra to study geometry. It is closely related to differential topology and PDEs (geometric analysis).

32835 questions
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What kind of manifold is one with a trivial tangent bundle?

Possible Duplicate: Which manifolds are parallelizable? This question is all about vector bundles isomorphic to trivial ones (for the sake of completeness I've collected definitions I refer to below). Let us agree to call trivial those bundles…
Giuseppe Negro
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On Tu's definition of tangent space at a point of $\mathbb{R}^n$

In Tu's book, the tangent vector at $p \in U \subseteq \mathbb{R}^n$ is defined as an arrow emanating from $p$. However, when passing to manifolds, such an approach cannot be used, since we should fix a chart $(U,\phi)$ about $p$ and, next, to…
TheWanderer
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showing that a local diffeomorphism is a local isometry using first fundamental form

In differential geometry, there is a theorem about 1st fundamental form : A local diffeomorphism $f:S_1 \rightarrow S_2$ is a local isometry $\Leftrightarrow$ For any patch $\sigma$ of $S_1$, $\sigma$ and $f \circ \sigma$ have the same 1st…
NNNN
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Prove the diff Bianchi identity $d^\nabla R^\nabla=0$ using that $(d^\nabla\circ d^\nabla)\circ d^\nabla = d^\nabla\circ (d^\nabla\circ d^\nabla)$

I found on page 25 of Arthur Besse’s “Einstein Manifolds” that the differential Bianchi identity $$d^\nabla R^\nabla=0$$ follows from the alternate definition of the Riemann curvature tensor as the second covariant exterior…
Rodrigo
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Problem about finding Gaussian curvature

The problem is about Gaussian Curvature When we define $M=r(u,v)$ be the surface parametrized by $(u,v)$ in $\mathbb{R}^3$ and $N$ be a unit normal vector of $M$. In my lecture we denote Gaussian curvature $\kappa=\det(B)/\det(A)$ where…
user76608
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Proving $d^\nabla( d^\nabla \omega) = F^\nabla \wedge \omega$

I'm going over the exterior covariant derivative $$d^\nabla : \Omega^k(E) \to \Omega^{k+1}(E)$$ of a vector bundle $E \to M$ and a connection $\nabla$ on $E$. There are some peculiar notions which I'm facing. I'm trying to verify that $$d^\nabla(…
Tepes
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Bundle isomorphism $E \to F$ from isomorphisms on each fiber $E_p \to F_p$?

If I have two real/complex vector bundles $E$ and $F$ over a smooth manifold $M$ and I can exhibit an isomorphism for each fiber $E_p \to F_p$ does this imply that there exists a bundle isomorphism $E \to F$?
Danlo
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For a 3D scalar field, does the Frenet-Serret frame always align with the principal curvatures of its isosurfaces?

In a given 3D scalar field that is smooth (in this case, with continuous 1st and 2nd derivatives, i.e. $f ∈ C^2(U)$), its gradient vector field is dual to the infinite set of isosurfaces in the function in the following sense: the gradient vector at…
twilsonco
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Understanding John Lee's Intuition Regarding Failure of Exactness of Covector Fields in the Punctured Plane

I am trying to understand what is meant by the following wording on page 297 of John Lee's Introduction to Smooth Manifolds, the 2nd Edition: "The key to constructing a potential function in Theorem 11.49 is that we can reach every point $x \in M$…
user480172
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Conidtions for a vector field to be geodesic?

I have a problem in which I have a vector field $u$ on a smooth manifold and want to find a Riemannian metric for which the vector field is geodesic, i.e. $$\nabla_u u = \sigma u,$$ and $\sigma$ is some smooth function. As a starting point, I would…
e4f5
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Why does the matrix of the bundle homomorphism $g:TM\to T^{*}M$ have entries $(g_{ij})$ and not $g_{ij}E_{k}^{i}$?

I am trying to understand the following statement from Lee's book introduction so smooth manifolds: Why does the matrix of $g:TM\to T^{*}M$ have entries $(g_{ij})$ and not $g_{ij}E_{k}^{i}$? As far as I am concerned I have the following: Let…
Hans
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(Non-)holonomic geodesic i.e. constrained geodesics

Fellows, I have been working with classical mechanics during engineering classes to know Euler-Lagrange equation (EL) $\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0$. Its version is constrained by…
Bruno Peixoto
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What is the barycenter of a closed Riemannian surface?

Let $\Sigma$ be a closed Riemannian surface. I read a paper which said: We let $\Sigma_k$ denote the family of formal sums $$ \Sigma_k=\sum_{i=1}^k t_i \delta_{x_i} ; \quad t_i \geq 0, \quad \sum_{i=1}^k t_i=1 ; \quad x_i \in \Sigma, $$ endowed with…
Elio Li
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Coefficients of the first fundamental form.

I have the following exercise: Let $f:\mathbb{R}^3\rightarrow \mathbb{R}$, and let $S$ be a surface. Let $p$ be a regular value of $f$, and $S=f^{-1}(p)$. Determine the coefficients of the first fundamental form. I think that since I have to work…
user2699
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Lie bracket is 0, yet flows to do not commute.

If $X, Y$ are two smooth vector fields on a smooth manifold $M$, we have that $[X,Y]=0$ implies that the flows $\phi^X_s \circ \phi^Y_t = \phi^Y_t \circ \phi^X_s$ commute, wherever both sides are defined. However, I'm troubled by a "counterexample"…
Mark
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